Sodium Carbonate cannot be used in place of $$(NH_4)_2CO_3$$ for the identification of $$Ca^{2+}$$, $$Ba^{2+}$$ and $$Sr^{2+}$$ ions (in group V) during mixture analysis because :

In qualitative inorganic analysis for group V cations, which include calcium ($$Ca^{2+}$$), barium ($$Ba^{2+}$$), and strontium ($$Sr^{2+}$$), we precipitate them as carbonates using ammonium carbonate ($$(NH_4)_2CO_3$$). The question asks why sodium carbonate ($$Na_2CO_3$$) cannot be used instead.

First, consider the precipitation reaction. The carbonate ion ($$CO_3^{2-}$$) reacts with these metal ions to form insoluble carbonates:

$$ Ca^{2+} + CO_3^{2-} \rightarrow CaCO_3 \downarrow $$

$$ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \downarrow $$

$$ Sr^{2+} + CO_3^{2-} \rightarrow SrCO_3 \downarrow $$

However, if we use sodium carbonate ($$Na_2CO_3$$), it dissociates completely in water:

$$ Na_2CO_3 \rightarrow 2Na^+ + CO_3^{2-} $$

This provides a high concentration of carbonate ions ($$CO_3^{2-}$$). Magnesium ion ($$Mg^{2+}$$), which belongs to group IV, is not supposed to precipitate in group V. But magnesium carbonate ($$MgCO_3$$) is sparingly soluble and can precipitate if the carbonate ion concentration is sufficiently high:

$$ Mg^{2+} + CO_3^{2-} \rightarrow MgCO_3 \downarrow $$

Ammonium carbonate ($$(NH_4)_2CO_3$$) is used instead because it contains ammonium ions ($$NH_4^+$$). These ions suppress the carbonate ion concentration by reacting with it:

$$ CO_3^{2-} + NH_4^+ \rightarrow HCO_3^- + NH_3 $$

This reaction reduces the concentration of $$CO_3^{2-}$$ ions, making it insufficient to precipitate $$MgCO_3$$ (since $$MgCO_3$$ has a higher solubility product compared to the carbonates of $$Ca^{2+}$$, $$Ba^{2+}$$, and $$Sr^{2+}$$). Thus, $$Mg^{2+}$$ remains in solution and is not precipitated in group V.

In contrast, sodium carbonate does not provide any suppressing ions. The sodium ions ($$Na^+$$) do not react with carbonate ions and do not reduce their concentration. Therefore, the high $$CO_3^{2-}$$ concentration from $$Na_2CO_3$$ causes $$Mg^{2+}$$ to precipitate as $$MgCO_3$$ along with the group V cations.

Now, evaluating the options:

Option A states that $$Mg^{2+}$$ ions will also be precipitated. This aligns with our reasoning, as the high carbonate concentration from sodium carbonate leads to unwanted precipitation of magnesium carbonate.

Option B claims the concentration of $$CO_3^{2-}$$ ions is very low. This is incorrect because sodium carbonate provides a high concentration of $$CO_3^{2-}$$ ions due to complete dissociation.

Option C suggests sodium ions react with acid radicals. Sodium ions ($$Na^+$$) are inert and do not react with acid radicals in a way that affects the precipitation of group V cations.

Option D says $$Na^+$$ ions interfere with the detection of $$Ca^{2+}$$, $$Ba^{2+}$$, $$Sr^{2+}$$. Sodium ions do not form insoluble compounds and do not interfere directly; the issue is the precipitation of $$Mg^{2+}$$.

Hence, the correct answer is Option A.